As requested by Nilotpal Kanti Sinha in the comments, here's the code I used to check maximal prime factor occurrences for all primes up to $4\cdot10^8$.
This is written in Sage, which is basically Python 2 with built-in maths. Hopefully the functions next_prime(), previous_prime(), prime_divisors() and max() are all self-explanatory.
The approach is to test successive multiples of each prime to see if they are the maximal prime factor in the relevant prime gap.
def get_max_prime(n):
# Find the maximal prime factor in the prime gap containing n
pp = previous_prime(n)
np = next_prime(n)
fs = set([]) # Set of all prime factors in the gap
for c in range(pp+1, np):
for p in prime_divisors(c):
fs.add(p)
return max(fs)
# target and step for tracking progress
target = 10**6
step = 10**6
p = 3 # The prime to be tested
high = 0 # Tracks the deepest search
while True:
q = p # q will be a multiple of p
m = 0 # Will contain the maximal prime factor in a gap
c = 1 # Multiplier
while(m != p):
c = c + 1
q = p * c
m = get_max_prime(q)
if c > high: # Display new deepest search
print p,c
high = c
if p > target: # Display progress
print "Reached", target
target = target + step
p = next_prime(p)
I will prove that the limit exists and is precisely the Golomb-Dickman constant $\lambda$. First, let me sketch how one can show that the expected value of $\frac{\log(P_1(n))}{\log(n)}$, where $P_1(n)$ denotes the largest prime factor of $n$, is precisely $\lambda\approx0.62433$ via probability theory.
We say that an integer $n\geq1$ is $y$-smooth if every prime divisor of $n$ is $\leq y$. We also say that $n\geq1$ is $y$-powersmooth if every prime power divisor of $n$ is $\leq y$. Let's denote by $S(x,y)$ and $PS(x,y)$ the set of $y$-smooth and $y$-powersmooth numbers $\leq x$ respectively. It is a theorem of Dickman that
$$F_S(\nu):=\lim_{x\to+\infty}\frac{1}{x}|S(x,x^{\nu})|=\rho(1/\nu)$$
where $\rho(t)$ is Dickman's function satisfying the differential equation $t\rho'(t)+\rho(t-1)=0$.
For large $x\gg0$, choose a random $tx\in[0,x]$ where $t\in[0,1]$ (that is, consider the random variable $xT\sim U(0,x)$ where $U(0,x)$ is the discrete uniform distribution that "looks" continuous for large $x$). What is the probability that $\frac{\log(P_1(tx))}{\log(tx)}\leq\nu$? Well
$$tx\in S(x,x^\nu)\iff P_1(tx)\leq x^\nu\iff \frac{\log(P_1(tx))}{\log(tx)}\leq\frac{\nu\log(x)}{\log(tx)}=\frac{\nu}{1+\frac{\log(t)}{\log(x)}}\sim\nu$$
so, for large $x\gg0$, we have $P\left(\frac{\log(P_1(xT))}{\log(xT)}\leq\nu\right)\sim P(xT\in S(x,x^\nu))=\frac{1}{x}|S(x,x^\nu)|\sim F_S(\nu)$. This means that $F_S(\nu)$ is the cumulative distribution function of $\frac{\log(P_1(n))}{\log(n)}$ so we can compute the expected value as
$$\lim_{x\to+\infty}\frac{1}{x}\sum_{n\leq x}\frac{\log(P_1(n))}{\log(n)}=\int_0^1\nu F_S'(\nu)d\nu=\int_0^1\frac{-\rho'(1/\nu)}{\nu}d\nu$$
$$=\int_0^1\rho\left(\frac{1}{\nu}-1\right)d\nu=\int_0^\infty\frac{\rho(t)}{(1+t)^2}dt=\lambda$$
Now we denote the greatest prime power divisor of $n\geq1$ as $Q_1(n)$. For example,
$$Q_1(24)=Q_1(2^3\times3)=2^3=8\text{ and }Q_1(72)=Q_1(2^3\times3^2)=3^2=9$$
We want to compute the expected value of $\frac{\log(Q_1(n))}{\log(n)}$ is a similar manner as before. The key idea is to note that, for large $x\gg0$ and $tx\in[0,x]$ chosen uniformly, we have
$$tx\in PS(x,x^\nu)\iff Q_1(tx)\leq x^\nu\iff\frac{\log(Q_1(tx))}{\log(tx)}\leq\frac{\nu\log(x)}{\log(tx)}\sim\nu$$
so, again, $P\left(\frac{\log(Q_1(xT))}{\log(xT)}\leq\nu\right)\sim\frac{1}{x}|PS(x,x^\nu)|$. Thus, if we had $\lim\limits_{x\to+\infty}\frac{|S(x,x^\nu)|-|PS(x,x^\nu)|}{x}=0$ (which I'll prove in just a moment), it would immediately follow that
$$\lim_{x\to+\infty}\frac{1}{x}|PS(x,x^\nu)|=\lim_{x\to+\infty}\frac{1}{x}|S(x,x^\nu)|=F_s(\nu)=\rho(1/\nu)$$
$$\Rightarrow\lim_{x\to+\infty}\sum_{n\leq x}\frac{\log(Q_1(n))}{\log(n)}=\int_0^1\nu F_S'(\nu)d\nu=\int_0^\infty\frac{\rho(t)}{(1+t)^2}dt=\lambda$$
as desired. So it suffices to show that $\lim\limits_{x\to+\infty}\frac{|S(x,x^\nu)|-|PS(x,x^\nu)|}{x}=0$.
First observe that every $y$-smooth number is $y$-powersmooth. Thus we have that
$$0\leq|S(x,y)\setminus PS(x,y)|=|S(x,y)|-|PS(x,y)|$$
For every prime $p\leq y$, denote $M_p(x,y)=\{p^{1+\lfloor\log_p y\rfloor}k\leq x\}$ the set of multiples of $p^{1+\lfloor\log_p y\rfloor}$ which are $\leq x$ where $p^{1+\lfloor\log_p y\rfloor}$ is the least power of $p$ that exceeds $y$.
Now observe that, if $n\in S(x,y)\setminus PS(x,y)$, then $n$ is a product of powers of primes $\leq y$ as $n=\prod_{p_i\leq y}p_i^{\epsilon_i}$ where $\epsilon_i$ can be zero and, at the same time, $\exists p_i\leq y$ such that $p_i^{\epsilon}>y\Rightarrow \epsilon_i\geq1+\lfloor\log_{p_i}y\rfloor$. But this means that $n\in S(x,y)\setminus PS(x,y)$ implies that $\exists p\leq y$ prime such that $n\in M_p(x,y)$ so $S(x,y)\setminus PS(x,y)\subseteq\bigcup_{p\leq y}M_p(x,y)$. Thus, we can bound $|S(x,y)|-|PS(x,y)|$ by
$$|S(x,y)|-|PS(x,y)|=|S(x,y)\setminus PS(x,y)|\leq\left|\bigcup_{p\leq y}M_p(x,y)\right|\leq\sum_{p\leq y}|M_p(x,y)|$$
$$=\sum_{p\leq y}\left\lfloor\frac{x}{p^{1+\lfloor\log_p y\rfloor}}\right\rfloor\leq\sum_{p\leq y}\frac{x}{p^{1+\lfloor\log_p y\rfloor}}\leq\sum_{p\leq y}\frac{x}{p^{\log_p(y)}}=\frac{x}{y}\sum_{p\leq y}1=\frac{\pi(y)}{y}x$$
$$\Rightarrow\boxed{\therefore 0\leq\frac{|S(x,y)|-|PS(x,y)|}{x}\leq\frac{\pi(y)}{y}}$$
Finally, note that $\frac{\pi(y)}{y}\sim\frac{1}{\log(y)}\to0$ by the prime number theorem which, in particular, implies $\lim\limits_{x\to+\infty}\frac{|S(x,x^\nu)|-|PS(x,x^\nu)|}{x}=0$. In fact, we only need $\frac{\pi(y)}{y}\to 0$ (i.e. the zero natural density of prime numbers) which is not that hard to prove.$\ \square$
Best Answer
As stated in lulu's comment, we know that $x,y,z$ are a Pythagorean triple and that we require $$x=2mn\quad y=m^2+n^2\quad z=m^2-n^2\quad m=n+1$$
By Størmer's theorem and https://oeis.org/A117581 we know that the largest consecutive $17$-smooth integers are $336140$ and $336141$. Testing all pairs of consecutive $17$-smooth integers up to this limit produces the largest answer $$12495000^2=12495001^2-4999^2$$ so the original observation is true.
Raw output from program for consecutive $17$-smooth integers $\ge2499$ :